Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am not very experienced in mathematical notation, so please excuse some terminology misuse or formatting shortcomings.

I have a project in which a value needs to increase from a set minimum to a set maximum in a set number of seconds. It is easy to calculate the linear value based on ratios.

Let $v$ = desired value, $n$ = minimum limit, $x$ = maximum limit, $t$ = elapsed time, and $t_x$ = allocated time:

$$v = \frac{t}{t_x}(x-n) + n.$$

Thus if my values are:

$$n = 5, x = 90, t_x = 1800 \text{ (half hour)}$$

For elapsed time of $5$ minutes ($600$ s):

$$v = \frac{600}{1800} (90-5) + 5 = 33.3.$$

The problem is I want to change this linear growth to exponential growth, and I'm not sure how to alter the formula.

So instead of $33.3$ at $5$ minutes, I would rather have $13$ for example. (Slow initial change, rapid later change.)

How can I insert an exponential growth factor into my equation and honor the minimum and maximum values allowed?

share|cite|improve this question
up vote 3 down vote accepted

I will change notation slightly. Our initial smallest value is $a$, and our largest value, after say $k$ seconds, is $b$. So every second our amount gets multiplied by $(b/a)^{1/k}$, the $k$-th root of $b/a$. At elapsed time $t$ seconds, where $0 \le t \le k$, the value is $$a \left(\frac{b}{a}\right)^\frac{t}{k}.$$

This is what would happen if we have an initial population $a$ of bacteria, growing under ideal conditions, and ending up with population $b$ after $k$ seconds. The formula above gives the population at time $t$, where $0 \le t \le k$.

It is also what happens if we have an initial amount of money $a$, which under continuous compounding grows to $b$ in time $k$.

Remark: The quantity $Q$ grows exponentially if and only if the quantity $\log Q$ grows linearly. So alternately, you could translate your knowledge about linear growth to a formula about exponential growth.

share|cite|improve this answer
Your explanation was extremely helpful as well as a clear presentation of the formula. I admit I am more encouraged to return to studying mathematics! – JYelton Aug 1 '12 at 17:28

Let your model be $v(t) = v_0 e^{\alpha t}$, where $v_0$ and $\alpha$ are constants to be determined.

From your data, you want $v(0) = n$, and $v(t_x) = x$.

This immediately gives $v_0 = n$, and then we have $v(t_x) = n e^{\alpha t_x}$, from which we get $\alpha = \frac{1}{t_x} \ln \frac{v(t_x)}{n} = \frac{1}{t_x} \ln \frac{x}{n}$.

So the model is $$ v(t) = n e ^ { \frac{t}{t_x} \ln \frac {x}{n}}.$$

share|cite|improve this answer

I see several sorts of ways to proceed. So let's suppose we had minimum $5$, max $90$, and $1800$ seconds to get there. Then we might have an exponential of the form $Pe^{rt}$ or perhaps $Pe^{rt} + c$. We might choose $f(t) = e^{rt} + 4$, so that $f(0) = 5$ (the minimum). We ask ourselves what $r$ would make it so that $f(1800) = 90$, the max?

Then we'd look at $e^{r1800} = 90 - 4 = 86$, or $\ln 86/1800 = r$ (that's really small). This would give the start of the exponential, growing slowly and then moreso.

But every choice was arbitrary, so let's look at a few other things you might do. Perhaps you want to change the rate of growth. You might do $5e^{rt}$, finding the correct $r$ again. Or you might do something like $Pe^{r(t - 10)}$, shifting along the exponential curve to places that change more slowly, etc.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.